Question: The equation of a circle $C$ is $x^2+y^2-12x-2y+36 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-12x) + (y^2-2y) = -36$ $(x^2-12x+36) + (y^2-2y+1) = -36 + 36 + 1$ $(x-6)^{2} + (y-1)^{2} = 1 = 1^2$ Thus, $(h, k) = (6, 1)$ and $r = 1$.